Question

QUESTION 4 The diagram below shows a 1 600 x 600 mm precast concrete portal culvert under a freeway The culvert is laid at a slope of 1,8 %) and has wingwalls at 45° which gives an entrance los coefficient of 0,2. The water surface elevations upstream and downstream are 23.928 m ang 23.085m respectively as shown. The invert level on centreline is 22.235 m as shown. Manning's roughness coefficient (n) for concrete 0,012 X Г 23.928m 23 085m 22.235m 18.500m h 20.500m a L X +200 Cad Soffit 600 Section X-X Calculate the invert and soffit levels at "a" 4.1 (2) 4.2. Calculate the headwater depth (HW) above the point "b". (3) 4.3. Will this culvert be flowing under inlet or outlet control? (3) 4.4 Determine the discharge (Q) in m3/s through the culvert. (6) 4.5 Determine the outlet velocity for the conditions given. (2)

Answer 1

23.928 m 23. 0857 Hero L k 18.500m _ 20.500 m . . 22.235m E ke -0.2 Slope =1.87 = 0.018 Invert level on center line = 22.235 m n -0.012 Head water level (HHW ) = 23.928 - 22.565 = 1.363 m Height of cubiert - Goomm =06 m HHW > Height of culvert .. It's Inlet controlled condition 49. Inlet level ca 10.018 X20.500)+22:235 5 22.235 -0.369 = 21.866 m 1 soffit level = 0.369 m. 92 already above HHW - 1.363m n.

How > Height of culvert 1:363 10.6 =) 7. Inler controlled condition 4.45 Discharge Av = (1.240.635 (2X0:6 012 211.240.6))(0.018.)? LxB) [+ RM3 Skz] = (1.240.6.) x 60m2(0.23 0.018, " = 2:753 m/s 4:59 cuplet velocity = $ 2"s = 6.biz) 10.25"3 (0.016) 2/ 22 K = 3.8236 m/sec ll.